The reaction involved here is the Tollen's test, which detects aldehydes and alpha-hydroxy ketones, as they are able to reduce the ammoniacal silver nitrate solution to metallic silver, producing a mirror-like coating on the glassware. This test specifically distinguishes these reducing sugars from non-reducing sugars and is often used to identify the presence of an aldehyde functional group.

Let's analyze each option:

A. Formic acid - Formic acid ($HCOOH$) is a carboxylic acid but importantly, it can also reduce silver ions to metallic silver because it can act as a reducing agent just like aldehydes. Thus, it will give a positive Tollen's test.

B. Formaldehyde ($HCHO$) is the simplest aldehyde, and aldehydes are known to give positive results with the Tollen's test since they can be oxidized to carboxylic acids, reducing the silver ions in the process to metallic silver.

C. Benzaldehyde ($C_6H_5CHO$) is an aromatic aldehyde and, similar to formaldehyde, it can be oxidized to an acid, thereby reducing the silver ions to silver in the Tollen's test, giving a positive result.

D. Acetone ($CH_3COCH_3$) is a ketone. Generally, ketones do not undergo oxidation easily and do not reduce ammoniacal silver nitrate, and thus, do not give a silver mirror with the Tollen's reagent.

Therefore, the compounds that will give a silver mirror with ammoniacal silver nitrate (positive Tollen's test) are Formic acid, Formaldehyde, and Benzaldehyde, but not Acetone.

The correct answer from the options given is therefore:

Option C: A, B, and C only.